Computing Matrix, finding kernel and image

by specialnlovin
Tags: bases, image, kernel, linear algebra
 P: 19 Let T: R[x]2$$\rightarrow$$ R[x]3 be defined by T(P(x))=xP(x). Compute the matrix of x with respect to bases {1,x,x2} and {1,x,x2,x3}. Find the kernel and image of T. I know how to do this when given bases without exponents, however I do not know exactly what this is saying and therefore am having a hard time starting it.
 Mentor P: 16,570 Let's compute the first column. You take the first basis element, that is 1. Now you'll have to express T(1)=x in terms of the basis {1,x,x²,x³}. That would be (0,1,0,0). So the first column would consist out of $$\left(\begin{array}{ccc} 0 & ? & ? \\ 1 & ? & ? \\ 0 & ? & ? \\ 0 & ? & ? \end{array} \right)$$ Now for the second and third column, you'll have to express T(x) and T(x²) in terms of the basis {1,x,x²,x³}.
 P: 19 Okay so matrix T(x)=x2 and with respect to the basis {1,x,x²,x³} the second column would be (0,0,1,0), T(x2)=x3 and with respect to the basis {1,x,x²,x³} would be (0,0,0,1). I would then say that T=(0 0 0) (1 0 0) (0 1 0) (0 0 1) with respect to {1,x,x²,x³} right? Not with respect to {1,x,x²} and {1,x,x²,x³} so then to find the im(T) I just use (0 0 0) (1 0 0) (0 1 0) (0 0 1) and solve it with respect to the basis {1,x,x²,x³} right? Then how would I go about solving for the ker(T)?
Mentor
P: 16,570

Computing Matrix, finding kernel and image

For the Ker(T). Take your matrix A. Then wonder when Ax=0. This you should be able to solve easily (it's a system with 3 unknowns and 4 equations)
 Mentor P: 16,570 So, if (x,y,z) is in the kernel, then you must have $$\left( \begin{array}{ccc} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x\\ y\\ z \end{array} \right) = 0$$ It's not hard to see that this can only be the case iff x=y=z=0
 P: 19 right, that just seemed way too easy I thought I was doing it wrong. Then the im(T) is the span of e2, e3, and e4
 Mentor P: 16,570 Yes, your image is correct to. And remember, math doesnt have to be difficult

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